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Mathematics Revision Guides – Circle Theorems Page 1 of 28
Author: Mark Kudlowski
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Mathematics Revision Guides
Level: GCSE Higher Tier
CIRCLE THEOREMS
Version: 3.5 Date: 13-11-2016
Mathematics Revision Guides – Circle Theorems Page 2 of 28
Author: Mark Kudlowski
CIRCLE THEOREMS
Recall the following definitions relating to circles:
A circle is the set of points at a fixed distance from the centre.
The perimeter of a circle is the circumference, and any section of it is an arc.
A line from the centre to the circumference is a radius (plural: radii).
A line dividing a circle into two parts is a chord.
If the chord passes through the centre, then it is a diameter.
A diameter divides a circle into two equal parts, and its length is twice that of a radius.
If the chord is not a diameter, the two resulting unequal regions of the circle are segments. The minor
segment is the smaller; the major segment the larger.
A sector is a region of a circle bounded by two radii. The smaller one is the minor sector, the larger
one the major sector.
A tangent is a line touching a circle at one point.
Mathematics Revision Guides – Circle Theorems Page 3 of 28
Author: Mark Kudlowski
The angle at the circumference subtended by a
diameter is a right angle, or more simply, the
angle in a semicircle is a right angle.
The line AB is a diameter of the circle, passing
through the centre, O. Angle AXB is therefore a
right angle.
Proof.
The triangle AXB can be divided into
two smaller triangles, AOX and BOX.
Both triangles are isosceles because
OA, OX and OB are all radii and
therefore equal in length.
AXB = + .
OAX and OXA (marked ) are
therefore equal,
and AOX is thus (180-2°.
Similarly OBX and OXB (marked
) are equal,
with BOX = (180-2°.
Since AOX and BOX form a
straight line, their angle sum is 180°.
Hence (180-2)° + (180-2)° = 180°
360-2-2)° = 180°
2+ 2 = 180°
+ = 90°.
AXB is a right angle.
Mathematics Revision Guides – Circle Theorems Page 4 of 28
Author: Mark Kudlowski
The angle subtended at the centre of a circle is
double the angle subtended at the circumference.
Points A, X and B lie on the circumference of a circle
centred on O.
The angle AOB at the centre is double the angle AXB at
the circumference (labelled ),
Proof.
The figure OAXB can be divided into two smaller
triangles, AOX and BOX. Both triangles are
isosceles because OA, OX and OB are all radii and
therefore equal in length.
AXB = + .
OAX and OXA (marked ) are therefore equal,
and AOX is thus 180-2 °.
Similarly OBX and OXB (marked ) are
equal, with BOX = 180-2°.
Since angles AOX, BOX and AOB add up to 360°
around O, it follows that
AOB = 360°- (180° – 2 ) – ( 180° – 2)
AOB = 360° - (360° - 2 - 2)
AOB = 2( + .
The angle AOB at the centre is double the angle AXB at the circumference .
Mathematics Revision Guides – Circle Theorems Page 5 of 28
Author: Mark Kudlowski
Although the ‘arrowhead’ configuration is the one most commonly shown in textbooks, there are other
patterns where the rule can be applied – in each case below, angle AOB at the centre is double the angle
AXB at the circumference .
In the left-hand diagram, AOB at the centre is reflex because AXB at the circumference is obtuse.
In the middle diagram, point B has moved such that XB is a diameter of the circle, and points X, O and
B are collinear.
In the right-hand diagram, point X has moved such that the line XB now crosses line OA.
Mathematics Revision Guides – Circle Theorems Page 6 of 28
Author: Mark Kudlowski
Angles subtended by the same chord are equal.
Let AB be a chord of a circle.
APB and AQB (marked ) are equal, as are
ARB and ASB (marked ) .
Also, the sum of the angles on opposite sides of
the chord AB is equal to 180°.
(See the section on Cyclic Quadrilaterals for
further details.)
This circle does not actually have the chords AB or
PQ drawn on it, but the rule still holds: APB and
AQB are equal, as are PAQ and PBQ.
Mathematics Revision Guides – Circle Theorems Page 7 of 28
Author: Mark Kudlowski
The bisector of a chord is a diameter.
Any line drawn across a circle is a chord,
and the perpendicular bisector of the chord
passes through the centre, and is therefore
also a diameter of the circle.
In the diagram, the perpendicular bisector of
chord AB is the line PQ. Since PQ passes
through the centre O, it is also a diameter.
Point X is the midpoint of the chord AB and
also lies on the diameter PQ.
The triangle AOB is isosceles because OA
and OB are radii, and hence equal.
Additionally triangles AOX and BOX are
congruent.
Notice also the symmetry of the triangle
AOB – it is isosceles because AO and OB are
both radii.
Mathematics Revision Guides – Circle Theorems Page 8 of 28
Author: Mark Kudlowski
A tangent and a radius meet at right angles; also, two tangents drawn from a point to a circle
describe two lines equal in length between the point and the circle.
The lines TX and TY both originate from the point T and form tangents with the circle at points X and Y.
These tangents are perpendicular to the radii of the circle OX and OY, and are also equal in length.
Because the triangles OXT and OYT also have side OT in common, both also have all three sides equal,
and are therefore congruent. As a consequence, angles XOT, YOT (labelled ) and angles XTO, YTO
(labelled ) are also equal.
Mathematics Revision Guides – Circle Theorems Page 9 of 28
Author: Mark Kudlowski
Angles in the alternate (opposite) segment are equal.
In the diagram below left, the chord BX and the tangent ST meet at X.
Let SXB be the angle between the chord and the tangent.
Let XAB be the angle in the alternate segment (bounded by chord BX).
angles SXB and XAB (shaded and labelled ) are equal.
There is another angle pair in the right-hand diagram.
This time, TXA is the angle between the chord AX and the tangent ST.
Let XBA be the angle in the alternate segment (bounded by chord AX)
Then angles TXA and XBA (shaded and labelled ) are equal.
Mathematics Revision Guides – Circle Theorems Page 10 of 28
Author: Mark Kudlowski
Proof. (Using angles at the centre)
The line ST is a tangent to the circle centred on O, and
is the angle between TX and the chord XA.
We begin by drawing radii at OX and OA.
Because the tangent ST and the radius OX meet at right angles,
OXS = OXT = 90°.
Hence OXA = (90 - )°.
Furthermore, OAX also = (90 -) °, since XOA is isosceles (OX and OA are radii).
The angles in a triangle add to 180°, and so XOA = (180 – (180 – 2))° = 2.
But, the angle at the centre (XOA) is double the angle at the circumference () ,
so XOA also = 2.
Angles and are equal.
Mathematics Revision Guides – Circle Theorems Page 11 of 28
Author: Mark Kudlowski
A cyclic quadrilateral is one that can be inscribed in a circle, in other words one whose vertices lie on
a circle’s circumference.
The opposite angles of a cyclic quadrilateral add up to 180°.
Thus, in the diagram shown right, angles and have a sum of
180°, as do angles and .
Proof (using angles at the centre)
Take the quadrilateral ABCD inscribed in the circle centred on
O.
By the rules of angles at the centre, if BAD = , then the
obtuse angle BOD = 2
Similarly, if BCD = , then the reflex angle BOD = 2
Since 2 + 2 = 360°, + = 180°.
The exterior angle of a cyclic quadrilateral is equal to the
interior opposite angle.
Let BAD = .
Since the opposite angles of a cyclic quadrilateral add to 180°,
BCD = 180-
Extending side DC to point P, BCP + BCD = 180° since
DCP is a straight line.
Hence BCP = 180° - (180-) = .
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Author: Mark Kudlowski
Examination questions usually feature more than just one of the theorems, and often include some
algebra and Pythagoras as well.
Also, diagrams in exam questions are not usually drawn accurately. This is also true for the examples
in this section.
Example (1): The points A , B, C and D lie on a circle with
centre O.
i) Given DAC = 26°, find angles DBC and ACD.
ii) Explain why angle BCD is not a right angle.
i) Angle DBC = 26° because it is in the same segment as
angle DAC.
As COA is a diameter of the circle, angle CDA = 90°
because the angle in a semicircle is a right angle.
Angle ACD = 180°- (90 + 26)° = 64°.
More simply, the two acute angles in a right-angled triangle
add to 90°, so angle ACD = (90 – 26)° = 64°.
ii) Angle BCD is not a right angle because BD is not a
diameter of the circle.
This is the converse of the “angle in a semicircle” theorem.
Mathematics Revision Guides – Circle Theorems Page 13 of 28
Author: Mark Kudlowski
Example (2): Points Q and S lie
on a circle centred on O.
SR is a tangent to the circle at S.
Angle QRS = 39° and angle SOQ
= 78°.
Prove that triangle SQR is
isosceles.
We know that SOQ is isosceles
because OQ and OS are radii, and
therefore equal.
This makes OSQ = OQS, and
as the angle sum of a triangle is
180°, OSQ = ½(180-78)° = 51°.
But we are also told that SR is a
tangent, so OSR (the angle
between the radius OS and the
tangent SR) is a right angle.
But, OSR =
OSQ + QSR = 90°,
so QSR = (90 – 51)° = 39°
QSR = QRS, and so QSR is isosceles.
Example (3): Which of the following quadrilaterals are cyclic ?
i) a rectangle; ii) a rhombus; iii) a square; iv) a kite with a pair of opposite right angles ?
A rectangle has all its angles equal to 90°, as does a square. Both are therefore cyclic, since opposite
pairs of angles add to 180°.
The kite with a pair of opposite right angles is also cyclic for the very same reason.
A rhombus has opposite pairs of angles equal, but because they are not generally right angles, a
rhombus is not cyclic.
Mathematics Revision Guides – Circle Theorems Page 14 of 28
Author: Mark Kudlowski
Example (4): Points P, Q, R and S lie on a circle.
PQ = QR, and angle PQR = 104°.
Explain why angle QSR = 38°.
If PQ = QR, then PQR is isosceles, and hence
QPR = ½(180 – 104)° = 38°.
Also QPR and QSR are in the same segment,
so QSR is also = 38° because angles in the same
segment are equal.
Mathematics Revision Guides – Circle Theorems Page 15 of 28
Author: Mark Kudlowski
Example (5):
Line TA is a tangent to the circle centred on O.
Angles BAC and BAT are 68° and 58° respectively.
Calculate angles BOC and OCA.
The angle at the centre is double the angle at the
circumference, so BOC = 2 68° = 136°.
Then we use the alternate segment theorem to deduce
that BCA = BAT = 58°.
This angle BCA is then divided into two smaller angles,
namely BCO and OCA..
BOC is isosceles, so BCO = ½(180-136)° = 22°.
Finally, angle OCA = BCA - BCO
= (58 – 22)° = 36°.
Mathematics Revision Guides – Circle Theorems Page 16 of 28
Author: Mark Kudlowski
Example (6): The quadrilateral ABCD is
cyclic, and PAQ is a tangent to the circle.
Also, BC = CD, QAB = 41°
and BAD = 82°.
Show that AD and BC are parallel, giving
reasons for justification.
If lines AD and BC were parallel, angles CBD
and ADB would form a pair of alternate angles
and be equal to each other.
Angle DCB = (180-82)° = 98° because opposite
angles of a cyclic quadrilateral sum to 180°.
Triangle DCB is isosceles,
so CBD = ½(180-98)° = 41°.
By the alternate segment theorem, ADB = 41°.
Angles CBD and ADB are indeed equal, so conversely lines AD and BC are parallel.
Mathematics Revision Guides – Circle Theorems Page 17 of 28
Author: Mark Kudlowski
Example (7): The circles centred on X and Y have the tangent AB in common and have radii of 9 cm
and 4 cm respectively.
i) Explain why ABYX is a trapezium.
ii) Show that AB = 12 cm.
i) The tangents at A and B meet their respective radii AX and BY at right angles, so angles XAB and
ABY are both equal to 90°, forming a pair of co-interior angles
Sides AX and BY are therefore parallel, i.e. ABYX is a trapezium.
ii) We label a point Q on the radius AX such that ABYQ is a rectangle.
Now XP = 9 cm and PY = 4 cm, so XY = 13 cm.
Also AX = 9 cm and BY = 4 cm, so QX = 9 – 4 = 5 cm.
The resulting triangle XQY has a hypotenuse of 13 cm and side QX = 5 cm.
We use Pythagoras to find the distance QY.
QY = 1214425169513 22 cm.
Because ABYQ is a rectangle, AB = QY.
Hence the distance AB = 12 cm.
Mathematics Revision Guides – Circle Theorems Page 18 of 28
Author: Mark Kudlowski
Example (8): The points A, B and C lie on the circumference of a circle.
The lines SAT and RBT are tangents to the circle at points A and B respectively.
These tangents meet at T.
Angle CAB = 46° and angle BTA = 54°.
Find angles BAT and ABC, justifying your answers.
Tangents originating from the same point are equal in length so TA = TB.
Triangle ATB is therefore isosceles, where angles BAT and ABT = ½(180-54) = 63°.
Additionally, ACB = BAT = 63° by the alternate segment theorem.
So ABC = (180 – (63 + 46))° = 71°.
Mathematics Revision Guides – Circle Theorems Page 19 of 28
Author: Mark Kudlowski
Example (9): A, B and C are points on
the circumference of a circle whose
centre is O.
TA and TB are tangents to the circle
Angle ATB = 44°.
Find angles x, y and z, giving reasons for
your answers.
Tangents originating from the same point are equal in length so TA = TB.
Triangle ATB is therefore isosceles, where angles BAT and ABT (labelled x) = ½(180-44) = 68°.
Also OBA = OAB (labelled y) because AOB is isosceles where OB and OA are radii. both radii),
Additionally, OBT = 90° because the tangent and radius at B meet at right angles.
We then find y by subtraction, i.e. (90-68)° = 22°.
The angle ACB (labelled z) is equal to angle x by the alternate segment theorem, i.e. it is also 68°.
We could have also calculated BOA = (180 – 44)° = 136°. The size of angle ACB (labelled z) is half
of that, or 68°, using the properties of angles at the centre.
Mathematics Revision Guides – Circle Theorems Page 20 of 28
Author: Mark Kudlowski
Example (10): Points A, B and C lie on the
circumference of a circle centred on O.
PQ is a tangent to this circle at P.
PCA = 56° and AOB = 94°.
Calculate angle OBC (labelled x), showing all working.
We use the alternate segment theorem to deduce that
ABC = 56°. This angle is further divided up into
angles ABO and OBC.
Since AOB is isosceles,
ABO = ½(180-94) = 43°.
Hence OBC (labelled x) = (56 – 43)° = 13°.
Mathematics Revision Guides – Circle Theorems Page 21 of 28
Author: Mark Kudlowski
Example (11): The circle below is centred on O, and points A, B and C lie on its circumference.
DCO is a straight line, DA is a tangent to the circle and ADO = 38°.
i) Calculate angle AOD.
ii) Calculate angle ABC, giving full reasons for your answer.
The tangent AD and the radius OA meet at 90° at point A, so OAD = 90°.
Hence AOD = (90 – 38)° = 52°
We then use properties of angles at the centre to deduce that the angle ABC at the circumference is half
the angle AOD at the circumference, so ABC is half of 52° or 26°.
Mathematics Revision Guides – Circle Theorems Page 22 of 28
Author: Mark Kudlowski
Example (12): The lines ABC and ADE
are tangents to the circle centred on O.
OD = 7 cm and AB = 24 cm.
F is the point where AO meets the
circumference.
Work out the distance AF.
Tangents AB and AD are equal in length,
as they originate from the same point A.
Therefore AD = 24 cm.
Also, ODA = 90° because the tangent
and radius at D meet at right angles.
Hence ODA is right-angled.
Using Pythagoras, we find that
OA = 2562549576724 22 cm.
The distance OF is 7 cm, being a radius of the circle, so AF = OA – AF = (25 – 7) cm = 18 cm.
Mathematics Revision Guides – Circle Theorems Page 23 of 28
Author: Mark Kudlowski
Example (13): In the circle below,
AOB = 63°.
The diameter through A and O is
extended to point D such that OC =
CD. Find angle x.
Because OCD is isosceles,
COD = x° and OCD = (180-2x)°.
Hence BCO = 2x° (180° in a straight
line), and, because OCB is isosceles
(OB , OC are radii) , it follows
OBC = 2x° as well.
Hence BOC = 180-4x°.
Since AOD is a straight line, the sum of angles AOB, BOC and COD must equal 180°.
This leads to the linear equation
63 + 180-4x + x = 180, so
63 - 3x = 0, angle x = 21°.
The angles of OCD are thus 21°,
21° and 138°, while those of BOC
are 42°, 42° and 96°.
Mathematics Revision Guides – Circle Theorems Page 24 of 28
Author: Mark Kudlowski
Example (14): Tie-in: Congruent triangles
Points A, B, R and P lie on a circle with centre O.
Points A, O, R and C lie on a different circle
The circles intersect at points A and R.
CPA, CRB and AOB are straight lines.
Prove that angle CAB = angle ABC.
Because AB is a diameter of the smaller circle,
ARB = 90° (angle in a semicircle).
CRB is a straight line, so ARC in the larger
circle = 90° (180° in a straight line)
.
Also, AOC = 90° (angles in the same segment
are equal) and BOC = 90° (180° in a straight
line).
Looking at triangles AOC and BOC we notice the
following facts:
Sides AO and OB are equal in length since both
are radii of the smaller circle.
Side OC is common to both triangles.
Angles AOC and BOC are equal.
The two triangles are therefore congruent
(SAS – two sides and the included angle)
Hence angles CAB and ACB (labelled )
are equal.
Mathematics Revision Guides – Circle Theorems Page 25 of 28
Author: Mark Kudlowski
An important result (IGCSE).
AB and CD are two chords which
intersect at point X, which is not
necessarily the centre of the circle.
Prove that the products (AX)(XB)
and (CX)(XD) are equal.
Proof.
Angles CAX and XDB (marked )
are equal because they are both
subtended by the segment CB.
Angles ACX and XBD (marked )
are equal because they are both
subtended by the segment AD.
Angles AXC and DXB (marked ) are
equal because they are vertically
opposite.
From the three results above, the triangles AXC and DXB are similar.
Sides CX and XB correspond, as they are opposite the angles marked .
The ratio of their lengths, XB
CX, is a constant value k.
Sides AX and XD correspond, as they are opposite the angles marked .
The ratio of their lengths, XD
AX, takes the same value k, as triangles AXC and DXB are similar.
Hence XB
CX=
XD
AX, and cross-multiplying, (AX)(XB) = (CX)(XD).
Example (15): In the circle shown, AX = 8 cm,
XB = 5 cm and CX = 10 cm. Find the length of
the chord CD.
Now, (AX)(XB) = 8 5 = 40, so (CX)(XD) = 40.
Therefore XD = 10
40= 4 cm.
Thus the length of chord CD is 10 + 4 = 14 cm.
This rule also holds when the chords intersect
outside the circle, as the following examples will
show.
Mathematics Revision Guides – Circle Theorems Page 26 of 28
Author: Mark Kudlowski
Example (16): In the circle shown, AB = 10 cm, XB = 5 cm and XD = 9 cm.
Find the length of the chord CD.
Now, AX = 10 + 8 = 18, so (AX)(XB) = 18 = 144. Since (CX)(XD) = 144, CX = 9
144= 16 cm.
Hence the chord CD is 16 – 9, or 7 cm, long.
Additionally, if the chord AB degenerates into a tangent at point A, the product (AX)(XB) becomes
(AX)(XA) or (AX)2 . In other words, we have (AX)
2 = (CX)(XD).
Example (17): In the circle shown, CD = 15 cm and XD = 9 cm.
Find the length of the tangent AX.
Here, CX = 15 + 5 = 20, and
(CX)(XD) = 20 = 100.
Since (AX)2 = 100, AX = 100 or
10 cm, i.e. the tangent AX is 10
cm long.
Mathematics Revision Guides – Circle Theorems Page 27 of 28
Author: Mark Kudlowski
Circles and triangles.
The circumcircle of a triangle is the circle passing through all of its vertices. Its centre is the
intersection of the perpendicular bisectors of its sides.
If the triangle is acute-angled, the
circumcentre will be inside the
triangle.
When we have an obtuse-angled triangle, the centre of the circumcircle lies outside the triangle, and
with a right-angled triangle, the circumcentre lies on the midpoint of the hypotenuse.
Mathematics Revision Guides – Circle Theorems Page 28 of 28
Author: Mark Kudlowski
The incircle of a triangle is the largest circle that can be inscribed within it. Its centre is the intersection
of the bisectors of its angles, and the three sides of the triangle are tangents to it.
Note that the tangents to the incircle do not generally coincide with the angle bisectors.
To complete the discussion, a line joining the
midpoint of a triangle’s side to the opposite
vertex is called a median. The three medians
of the triangle meet at a point called the
centroid of the triangle, which, incidentally,
is also the triangle’s centre of gravity.
The midpoints of sides AB, BC and CA are at
points P, Q and R.
The medians are AQ, BR and CP and they
intersect at the centroid X.